Verified Commit 1e5c9a8a authored by Anton Akhmerov's avatar Anton Akhmerov
Browse files
parents 1ea314f4 5afd3619
......@@ -27,6 +27,7 @@ nav:
- Semiconductors:
- Basic principles: '13_semiconductors.md'
- Doping and devices: '14_doping_and_devices.md'
- Summary and review: '15_SummaryAndReview.md'
- Attic:
- Magnetism: 'lecture_8.md'
- Fermi surface periodic table: 'fermi_surfaces.md'
......@@ -45,7 +46,7 @@ nav:
- Nearly free electron model: '11_nearly_free_electron_model_solutions.md'
- Band structures in 2D: '12_band_structures_in_higher_dimensions_solutions.md'
- Basic principles: '13_semiconductors_solutions.md'
# - Doping and devices: '14_doping_and_devices_solutions.md'
- Doping and devices: '14_doping_and_devices_solutions.md'
theme:
name: material
......
......@@ -72,7 +72,7 @@ dispersion2D()
Construct the Hamiltonian with basis vectors $(\pi/a,0)$ and $(-\pi/a,0)$, eigenvalues are
$$ E=\frac{\hbar^2}{2m} \left(\frac{\pi}{a}\right)^2 \pm \left|V_{10}\right|^2. $$
$$ E=\frac{\hbar^2}{2m} \left(\frac{\pi}{a}\right)^2 \pm \left|V_{10}\right|. $$
### Subquestion 2
......
......@@ -125,7 +125,7 @@ In semiconductors the Fermi level is between two bands. The unoccupied band is t
We can control the position of the Fermi level (or create additional excitations) making semiconductors conduct when needed.
Only the bottom of the conduction band has electrons and the top of the valence band has holes because the temperature is always smaller than the width of the band.
Only the bottom of the conduction band has electrons and the top of the valence band has holes because the temperature is always smaller than the size of the band gap.
Therefore we can approximate the dispersion relation of both bands as parabolic.
......@@ -227,7 +227,7 @@ $$n_h \approx N_V e^{-E_F/kT} = N_C e^{-(E_G-E_F)/kT} \approx n_e$$
Solving for $E_F$:
$$E_F = \frac{E_G}{2} - \frac{3}{4}kT\log(m_e/m_h)$$
$$E_F = \frac{E_C + E_V}{2} - \frac{3}{4}kT\log(m_e/m_h)$$
An extra observation: regardless of where $E_F$ is located, $n_e n_h = N_C N_V e^{-E_G/kT} \equiv n_i^2$.
......
......@@ -13,7 +13,7 @@ Electrons near the top of the valence band have a negative effective mass, their
### Subquestion 2
Holes near the top of the valence band have a positive effective mass, their energy decreases as $k$ increases from 0, and they have a negative group velocity for $k>0$.
Holes near the top of the valence band have a positive effective mass, their energy increases as $k$ increases from 0, and they have a negative group velocity for $k>0$.
### Subquestion 3
......@@ -106,18 +106,27 @@ This approximation indicates the chemical potential is "well bellow" the conduct
the valence band. This way, we only have a few electrons and a few holes per band, which allows us to
approximate Fermi statistics by Boltzmann's.
Assume $t_{cb}$ and $t_{vb}$ positive.
For electrons, do a taylor expansion around $k=0$ (min) of $E_{cb} = E_G - 2 t_{cb} [\cos(ka)-1],$.
For holes, do a taylor expansion around $k=0$ (max) of $E_{vb} = 2 t_{vb} [\cos(ka)-1]$.
Assume $t_{cb}$ and $t_{vb}$ positive.
Near $k=0$
$$ E_{cb} = E_G + t_{cb}(ka)^2$$
$$ E_{vb} = -t_{vb}(ka)^2 $$
### Subquestion 3
Now, $g_e$ and $g_h$ can be calculated.
??? hint "How?"
The approximated bands are quadratic! We've seen these before during the lecture.
### Subquestion 4
$$n_e = \int_{E_{cb}}^{E_{cb}+4t_{cb}} f(\varepsilon)g_c(\varepsilon)d\varepsilon$$
$$n_h = \int_{-4t_{vb}}^{0} (1-f(\varepsilon))g_h(\varepsilon)d\varepsilon$$
### Subquestion 5
For an intrinsic semiconductor the number of electrons excited into the conduction band must be equal
to the number of holes left behind in the valence band, so $p = n$. Impose this condition with the results of
......
......@@ -25,6 +25,18 @@ m_h, m_e = 1, .5
- Compute carrier density and Fermi level position of doped semiconductors
- Describe the functioning principles of semiconducting devices
??? info "Lecture video (Part 1/2)"
<iframe width="100%" height="315" src="https://www.youtube-nocookie.com/embed/rVdKd-0vAkk" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
??? info "Lecture video (Part 2/2)"
<iframe width="100%" height="315" src="https://www.youtube-nocookie.com/embed/RnUIMMbZqKY" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
??? info "Lecture video - Extra topics"
<iframe width="100%" height="315" src="https://www.youtube-nocookie.com/embed/1u305H4UiVs" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
## Adding an impurity to semiconductor
* Typical semiconductors are group IV (Si, Ge, GaAs).
......@@ -76,7 +88,7 @@ draw_classic_axes(ax, xlabeloffset=.2)
All donor/acceptor states at the same energy:
$$g_A(E) = N_A \delta(E-E_A),\quad g_D(E) = N_D \delta(E- E_D)$$
How large can $N_D/N_A$ be? The distance between donors should be such that the states don't overlap, so the distance must be much larger than 4 nm. Therefore **maximal** concentration of donors before donor band merges with conduction band is $N_D \lesssim (1Å/4\textrm{nm})^3 \sim 10^{-5}\ll N_C$.
How large can $N_D/N_A$ be? The distance between donors should be such that the states don't overlap, so the distance must be much larger than 4 nm. Therefore **maximal** concentration of donors before donor band merges with conduction band is $N_D \lesssim (1/4\textrm{nm})^3 \sim 10^{-5}\ll N_C$.
## Number of carriers
......
```python tags=["initialize"]
from matplotlib import pyplot as plt
import numpy as np
from math import pi
```
# Solutions for lecture 14 exercises
## Exercise 1: Crossover between extrinsic and intrinsic regimes
### Subquestion 1
Law of mass action:
$$ n_e n_h = \frac{1}{2} \left(\frac{k_BT}{\pi\hbar^2}\right)^3
(m_e^{\ast}m_h^{\ast})^{3/2}e^{-\beta E_{gap}}$$
Charge balance condition:
$$ n_e - n_h + n_D - n_A = N_D - N_A $$
### Subquestion 2
$$ n_{e} = \frac{1}{2}(\sqrt{D^2+4n_i^2}+D)$$
$$ n_{h} = \frac{1}{2}(\sqrt{D^2+4n_i^2}-D)$$
where $D = N_D - N_A$ and $n_i=n_{e,intrinsic}=n_{h,intrinsic}$.
Note that for N_D = N_A results for intrinsic semiconductors are recovered.
### Subquestion 3
If $D<<n_i$, then the doping is not important and results of intrinsic are
reproduced.
Contrarily, if $D>>n_i$, it's mostly the doping that determines $n_e$ and $n_h$.
The thermal factor becomes unimportant.
Check both cases with lecture notes approximated solutions by doing a Taylor expansion.
## Exercise 2: Donor ionization
### Subquestion 1
If all the dopants are ionized ($n_D=0$), the Fermi level gets shifted up towards the conduction band.
This result can be obtained when using results in Exercise 1 - Subquestion 2 and the following:
$$ n_e - n_h = N_D - N_A $$
??? hint "To find $E_F$"
Solve now for $E_F$ using $n_e$ and $n_h$ expressions obtained in the
[previous lecture](/13_semiconductors/#part-1-pristine-semiconductor).
### Subquestion 2
Now,
$$ n_D = N_D\frac{1}{e^({E_D-E_F})/k_BT+1} $$
### Subquestion 3
Because not all dopants are ionized, the charge conservation eq. becomes:
$$ n_e - n_h + n_D = N_D - N_A $$
Doing the same as in subquestion 1, an expression for $E_F$ can be found.
??? hint "Values"
For Germanium at $T=300K$, $n_i = 2.4 \times 10^{13} cm^{-3}$
This can be obtained when plugging $m_e^{\ast}=0.55 m_e$, $m_h^{\ast}=0.37 m_e$ and $T=300K$ in
[the results here](/13_semiconductors/#semiconductor-density-of-states-and-fermi-level).
## Exercise 3: Performance of a diode
### Subquestion 1
Intrinsic semiconductors have no impurities. Adding dopant atoms creates extra unbounded
electrons/holes depending on the n/p dopant atom added. Impurity eigenstates appear and
the $E_F$ level shifts (up/down for added donors/acceptors).
To make a diode a p-n junction is needed (extrinsic semiconductors). Drawing a diagram is
very helpful.
### Subquestion 2
Under reverse bias only two processes carry out current: electrons that may be thermally
excited into the conduction band (p-doped side) and holes that may be thermally
excited into the valence band (n-doped side).
### Subquestion 3
$$ I_s(T) \propto e^{-E_{gap}/k_BT}$$
## Exercise 4: Quantum well heterojunction in detail
### Subquestion 1
![](figures/diagram_14.png)
* Include the energy bands here. You can find them at the book's section 18.2
### Subquestion 2
This a "particle in a box" problem.
$$-\frac{\hbar^2}{2m_e^{\ast}} \nabla^2 \Psi_e = (E_e-E_c)\Psi_e $$
$$-\frac{\hbar^2}{2m_h^{\ast}} \nabla^2 \Psi_h = (E_v-E_h)\Psi_h $$
### Subquestion 3
$$E_e = E_c + \frac{\hbar^2}{2m_e^{\ast}} ((\frac{\pi n}{L})^2+k_x^2+k_y^2)}$$
$$E_h = E_v - \frac{\hbar^2}{2m_h^{\ast}} ((\frac{\pi n}{L})^2+k_x^2+k_y^2)}$$
### Subquestion 4
This is a 2D electron/hole gas. Apply 2D density of states (constant).
$$g_e = \frac{4 \pi m_e^{\ast}}{\hbar^2}$$
$$g_h = \frac{4 \pi m_h^{\ast}}{\hbar^2}$$
### Subquestion 5
L can be found here using previous subquestions.
Setting $$ E_e - E_h - E_c + E_v = 1 eV = \frac{\hbar^2}{2}(\frac{\pi n}{L}^2+k_x^2+k_y^2)
(\frac{1}{m_e^{\ast}}+\frac{1}{m_h^{\ast}})$$
By choosing the correct $n$, $k_x$ and $k_y$, L can be found as $6.85$ nm approx.
### Subquestion 6
For a laser one wants to fix the emission wavelength to a certain value. With
this setup the band gap is "easy" to design (set by L, which is fixed).
### Subquestion 7
If donor impurities are put outside of the well (on both sides, for example)
the donated electrons can reduce their energies by falling into the well,
but the ionized dopants remain behind. This gives an advanttage because an
extremely high mobility for electrons can be obtained within the quantum well
(there are no ionized dopants in the well to scatter off of).
This is called modulation doping.
```python tags=["initialize"]
from matplotlib import pyplot
import numpy as np
from scipy.optimize import curve_fit
from scipy.integrate import quad
from common import draw_classic_axes, configure_plotting
configure_plotting()
```
# Summary & review
??? info "Lecture video"
<iframe width="100%" height="315" src="https://www.youtube-nocookie.com/embed/NuQgJRQuJy8" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
......@@ -14,7 +14,7 @@ $$
We assume that in $d$ dimensions there are $d$ polarizations.
For 1D we have that $N = \frac{L}{2\pi}\int dk$, hence $g(\omega) = \frac{L}{2\pi v}$.
For 1D we have that $N = \frac{L}{2\pi}\int_{-k}^{k} dk$, hence $g(\omega) = \frac{L}{\pi v}$.
For 2D we have that $N = 2\left(\frac{L}{2\pi}\right)^2\int d^2k = 2\left(\frac{L}{2\pi}\right)^2\int 2\pi kdk$, hence $g(\omega) = \frac{L^2\omega}{\pi v^2}$.
......
Markdown is supported
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment